# Help with a Limit

I have tried to solve this limit with L'Hopital's rule; $\lim_{x\to0+}(1+x)^{lnx}$ but i applied several times with any result. I would appreciate if somebody can help me. Note: Wolfram alpha said that the result is 1, but it doesn't appear the steps to get it.

• I have merged Dr Sonnhard Graubner's answers. If you want to accept his answer, you will need to accept it again. – robjohn Oct 6 '14 at 17:47

hint: write $(1+x)^{\ln(x)}$ as $e^{\frac{\ln(1+x)}{\frac{1}{\ln(x)}}}$

• I did it, but i applied L'Hopital's rule several times, and the indeterminate doesn't disappear – egarro Oct 6 '14 at 17:04
• the first step is $\frac{\frac{1}{1+x}}{\frac{-1}{x\ln(x)^2}}$ and then $-\frac{x\ln(x)^2}{1+x}$ – Dr. Sonnhard Graubner Oct 6 '14 at 17:14
• And then we get the form 0*infinity, right? – egarro Oct 6 '14 at 17:22
• yes and then we write $-\frac{\ln(x)^2}{1+\frac{1}{x}}$ and the next step is the last one $\frac{2\ln(x)\frac{1}{x}}{-\frac{1}{x^2}}$ – Dr. Sonnhard Graubner Oct 6 '14 at 17:28
• Thank so much! I already get the result. – egarro Oct 6 '14 at 17:35

To apply l'Hopital's rule, you first need to get the something in the form $\lim_{x\to a}\frac{f(x)}{g(x)}$, where $f(x)$ and $g(x)\to 0$ as $x\to a$. In you case, if we let $L=\lim_{x\to0^+}(1+x)\ln x$, then taking $\ln$ of both sides, we get $$\ln L = \lim_{x\to 0^+}\ln x\ln (1+x)=\lim_{x\to 0^+}\frac{\ln (1+x)}{1/\ln x}$$ Now, you can apply l'Hopital's rule to the last expression, allowing you to find $\ln L$, then raise $e$ to that to find $L$.

• I did that, but i applied the rule like three times, with any result. – egarro Oct 6 '14 at 17:00

Without using Hospital rule also we can evaluate this limit.
Substitute $x=e^t$ then $x\to0^+$ is same as $t\to -\infty.$ $$\lim_{x\to0^+}(1+x)^{\ln x}=\lim_{t\to-\infty} (1+e^t)^t=\lim_{t\to-\infty} (1+O(e^t))=1.$$

• How did the $t$ (which goes to $-\infty$) disappear from the exponent? – robjohn Oct 6 '14 at 17:17
• @robjohn: We can use the binomial theorem. – Bumblebee Oct 6 '14 at 17:21
• With an exponent approaching $-\infty$? That seems hard to do. – robjohn Oct 6 '14 at 17:51

Hint: $f(x,y)=x^y$ is continuous at $(e,0)$ so \begin{align} \lim_{x\to0}\left[\left(1+x\right)^{1/x}\right]^{x\log(x)} &=\left[\lim_{x\to0}\left(1+x\right)^{1/x}\right]^{\lim\limits_{x\to0}x\log(x)}\\ \end{align} You can also write $x\log(x)=\dfrac{\log(x)}{1/x}$ and use L'Hospital once.

Alternate Approach \begin{align} \lim_{x\to0}\log\left(\left(1+x\right)^{\log(x)}\right) =\lim_{x\to0}\frac{\log(x)}{1/x}\lim_{x\to0}\frac{\log(1+x)}{x} \end{align} Each limit requires one application of L'Hospital.